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I need proof for the following question. Also, I want to know, can we apply the same for other conics. If yes, where and when... Please explain.

Show that there exists a point K on the major axis of E , having the property that for any chord $\overline{PQ}$ passing through K, $\dfrac{1}{PK^2} + \dfrac{1}{QK^2}$ is a constant. Also Show that $\lim_{e \to \infty}K = (2a, 0)$

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I have some problems reading this. Doesn't $e\to\infty$ mean that conic section will sooner rather than later become a hyperbola? Yes, the curve can gradually transform into a hyperbola (and eventually, as a limit case, a straight line), but for the answer to make any sense this gradual transformation must take place in such a way that e.g. the scale won't suddenly blow up. In other words, there has to be a normalizing condition that you want to keep as a secret. Not cool :-( IOW give us the equation of the curve, where $e$ is a parameter. – Jyrki Lahtonen Jul 19 '11 at 14:41

We use a brute force approach. Let our ellipse have equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ where $a \ge b >0$. Let $K=(p,0)$ be a point (if there is one) such that $1/(PK)^2+1/(QK)^2$ is the same for all chords $PQ$ through $K$.

We will, regrettably, cheat, by (sort of) looking up the answer. Suppose that there is such a magic point $K=(p,0)$. Consider the horizontal and vertical chords through $K$. For the horizontal chord, the two relevant squares of distances are $(a+p)^2$ and $(a-p)^2$. For the vertical chord, we calculate, by substituting $x=p$ in the equation of the ellipse. The two relevant squares of distances turn out to be each $(b^2/a^2)(a^2-p^2)$.

In order to satisfy our inverse square condition, we need $$\frac{1}{(a+p)^2}+\frac{1}{(a-p)^2}=\frac{2a^2}{b^2(a^2-p^2)}.$$ Solving for $p^2$ turns out to be easy: $$p^2=\frac{a^2(a^2-b^2)}{a^2+b^2}.$$

We have found the only possible candidates for $p$. Now we need to verify that that we get constant sum of inverse square distances for all chords through $(p,0)$.

The generic line passing through $(p,0)$ has equation of the form $y=m(x-p)$. It is somewhat more convenient to rewrite this as $ny=x-p$, where $n$ is the reciprocal of the slope. One line through $(p,0)$ gets missed. We hope it doesn't feel bad.

Let $(x_1,y_1)$ and $(x_2,y_2)$ be the two points where the line $ny=x-p$ meets the ellipse. We want, with $p$ as determined above, to show that $$\frac{1}{(x_1-p)^2+y_1^2}+ \frac{1}{(x_2-p)^2+y_2^2}$$ does not depend on $n$. Since $(x_i-p)^2=n^2y_i^2$, our condition reduces to showing that $$\frac{1}{n^2+1}\left(\frac{1}{y_1^2}+\frac{1}{y_2^2}\right)$$ is independent of $n$. Equivalently, we want to show that $$\frac{1}{y_1^2}+\frac{1}{y_2^2}\qquad\text{or equivalently}\qquad \frac{(y_1+y_2)^2-2y_1y_2}{y_1^2y_2^2}$$ is a constant times $n^2+1$.

So now we want (sort of) to find $y_1$ and $y_2$. The ellipse has equation $b^2x^2+a^2y^2=a^2b^2$. But since $ny=x-p$, by substituting, we arrive at the equation $$b^2(ny+p)^2 +a^2y^2=a^2b^2.$$ Temporarily, rewrite this as $Ay^2+By+C$. Since $y_1+y_2=-B/A$, and $y_1y_2=C/A$, we find that $$\frac{(y_1+y_2)^2-2y_1y_2}{y_1^2y_2^2} =\frac{B^2-2AC}{C^2}.$$

Now we need to compute. We have $$\frac{B^2-2AC}{C^2}=\frac{4b^4n^2p^2-2(b^2n^2+a^2)(b^2p^2-a^2b^2)}{(b^2p^2-a^2b^2)^2}.$$ The expression $a^2-p^2$ that occurs a couple of times in the above equation simplifies to $2a^2b^2/(a^2+b^2)$. The denominator is constant. After not much work, the numerator simplifies to $(n^2+1)(4b^4a^4)/(a^2+b^2)$, a constant times $n^2+1$, and we are finished.

At least temporarily. The parabola has obviously only one point of the above type, namely the vertex. (For fans of points at infinity, there is another one.) It would be interesting to look for a synthetic proof. There may be a simple one, since the expression for $p^2$ has obvious geometric significance.

Added: What happens when eccentricity becomes large.

Recall that $$p^2=\frac{a^2(a^2-b^2)}{a^2+b^2}.$$ Now fix $a$, and let the eccentricity get large. As the eccentricity gets large, $b^2$ approaches $0$. Note that $$p^2=a^2\frac{1-\frac{b^2}{a^2}}{1+\frac{b^2}{a^2}}$$ (we have divided top and bottom by $a^2$.)

It follows that $$\lim_{b^2\to 0} p^2=a^2.$$

Thus if we start with the standard ellipse, it is not true that $K$ approaches $(2a,0)$. All we can say is that $p$ is then awfully close to $(a,0)$ or $(-a,0)$, either end of the major axis. If the point $K$ moves smoothly as $b^2$ decreases, we can assert a bit more, that $K$ approaches $(a,0)$ or $(-a,0)$.

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Descartes would be pleased. I wonder, though, if there is some other (geometric?) proof. – leonbloy Jul 19 '11 at 20:12
@leonbloy: I expect, may think about it. If you look at the expression for $p^2$, you will see that it is related to the (square of) focal distance, with a little scaling. That gives a strong hint about the shape of a synthetic proof. – André Nicolas Jul 19 '11 at 20:18
Yes, brute force indeed. I haven't thought very deeply about this and maybe that's what you have in mind, but I think one reasonable way to look at the equation for $p^2$ would be to write $$\frac{p}{a} = \frac{\sqrt{a^2-b^2}}{\sqrt{a^2+b^2}}.$$ There should be a circle. Well, I don't quite see it, but that's where I would start. – t.b. Jul 20 '11 at 2:02
@Theo: Yes, a couple of shortcuts using symmetric functions, the rest crunch. Have some ideas about using classical ellipse stuff, not exactly high priority. – André Nicolas Jul 20 '11 at 3:41

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